As the titles says, I have to prove that $S^3\setminus S^1$ is connected.
I'm having a hard time solving this problem. The best idea I've come up with is to show somehow that, if $S^3\setminus S^1$ wasn't connected, then $S^1$ would be a non-trivial clopen in $S^3$, and that's a contradiction; to do this I've tried to use the fact that in these hypotheses $S^3\setminus S^1=A\cup B$ disjoint open sets and $S^3\setminus S^1=C \cup D$ disjoint closed sets, but I haven't been able to conclude anything (and I'm not sure this is the right way to proceed). Maybe I'm missing something trivial.
Any hints?
Best Answer
Using the stereographic projection, you can identify $S^3-\{point\}$ (the point not in $S^1$) to $\mathbb{R}^3$. The image of the circle by the stereographic projection is a circle if the center of the stereographic projection is not in the circle. If you remove a circle to $\mathbb{R}^3$ it is still connected.
This implies that $S^3-\{point\}\cup S^1$ is connected and $S^3-S^1$ is connected.